Three cards are dealt at random from a standard deck of 52 cards.  What is the probability that the first card is a 4, the second card is a $\clubsuit$, and the third card is a 2?
Explanation: There are 4 exclusive cases:

Case 1: first card not a $\clubsuit$ and second card not a 2.

There are 3 cards that are 4's but not a $\clubsuit$, so the probability for the first card is $\dfrac{3}{52}$. Next, there are 12 $\clubsuit$s remaining that aren't a 2, so the probability for the second card is $\dfrac{12}{51}$. Finally, there are four 2's remaining, so the probability for the third card is $\dfrac{4}{50}$.  Hence, this case gives a probability of $\dfrac{3}{52}\times \dfrac{12}{51}\times \dfrac{4}{50} = \dfrac{144}{132600}$.  (We leave the fraction in these terms rather than reducing because we know that we're going to need to add fractions later.)

Case 2: first card not a $\clubsuit$ and second card the 2$\clubsuit$.

There are 3 cards that are 4's but not a $\clubsuit$, so the probability for the first card is $\dfrac{3}{52}$. Next, there is only one 2$\clubsuit$, so the probability for the second card is $\dfrac{1}{51}$. Finally, there are three 2's remaining, so the probability for the third card is $\dfrac{3}{50}$.  Hence, this case gives a probability of $\dfrac{3}{52}\times \dfrac{1}{51}\times \dfrac{3}{50} = \dfrac{9}{132600}$.

Case 3: first card the 4$\clubsuit$ and second card not a 2.

There is only one 4$\clubsuit$, so the probability for the first card is $\dfrac{1}{52}$. Next, there are 11 $\clubsuit$s remaining that aren't a 2, so the probability for the second card is $\dfrac{11}{51}$. Finally, there are four 2's remaining, so the probability for the third card is $\dfrac{4}{50}$.  Hence, this case gives a probability of $\dfrac{1}{52}\times \dfrac{11}{51}\times \dfrac{4}{50} = \dfrac{44}{132600}$.

Case 4: first card the 4$\clubsuit$ and second card the 2$\clubsuit$.

There is only one 4$\clubsuit$, so the probability for the first card is $\dfrac{1}{52}$. Next, there is only one 2$\clubsuit$, so the probability for the second card is $\dfrac{1}{51}$. Finally, there are three 2's remaining, so the probability for the third card is $\dfrac{3}{50}$.  Hence, this case gives a probability of $\dfrac{1}{52}\times \dfrac{1}{51}\times \dfrac{3}{50} = \dfrac{3}{132600}$.

So the overall probability is $\dfrac{144+9+44+3}{132600} = \dfrac{200}{132600} = \boxed{\frac{1}{663}}$.